A Level Computer Science 9618 Paper 31, May/June 2026

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Question paper

9618/31 · May/June 2026

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1
a

A record data type is a type of composite user‑defined data type. Identify two other types of composite user‑defined data types. 1 ................................................................................................................................................ 2 ................................................................................................................................................

[2]
b(i)

A pointer data type is a non‑composite user‑defined data type. State the purpose of a pointer user‑defined data type.

[1]
b(ii)

Write the pseudocode statement to declare a pointer data type called Word to be used with STRING data.

[2]
2
a

Random file organisation can be used with a database management system. State one other example of where random file organisation could be used.

[1]
b

Describe how records are organised in random files.

[2]
c

Outline the process that would be used to find a record in a random file.

[3]
3

Numbers are stored in a computer system using binary floating‑point representation with: • 10 bits for the mantissa • 6 bits for the exponent • two’s complement form for both the mantissa and the exponent.

a

Calculate the normalised binary floating‑point representation of –10.8125 in this system. Show your working.

[3]
b

Calculate the denary value of the given normalised binary floating‑point number. Show your working. mantissa 1 0 1 0 0 0 0 1 1 1 exponent 0 0 0 1 1 1

[3]
4

Draw one line to connect each TCP/IP layer to its most appropriate description.

[4]
5

A virtual machine may be used to emulate computer hardware using software. Outline two benefits and two limitations of using a virtual machine. benefit 1 ........................................................................................................................................... .......................................................................................................................................................... benefit 2 ........................................................................................................................................... .......................................................................................................................................................... limitation 1 ........................................................................................................................................ .......................................................................................................................................................... limitation 2 ........................................................................................................................................ ..........................................................................................................................................................

[4]
6
Figure 1 of 1 for question 6
a

The diagram shows a logic circuit. Complete the truth table for the given logic circuit. Show your working.

[3]
b

Describe one benefit of using a Karnaugh map (K‑map) for solving logic problems.

[1]
c(i)

The Karnaugh map (K‑map) represents a logic circuit with four inputs. Draw loop(s) around appropriate group(s) in the K‑map to produce an optimal sum‑of‑products.

[3]
c(ii)

Write the Boolean expression from your answer to 6(c)(i) as a simplified sum‑of‑products. Do not carry out any further simplification.

[2]
7
a

Write this infix expression in Reverse Polish Notation (RPN). 21 * (15 – 3 + 8) / 7

[2]
b

Write this RPN expression in infix form. c d + a b – a c - * /

[3]
c

The RPN expression 15 9 – 24 3 / * 18 6 + / is to be evaluated. Show the changing contents of the stack as the RPN expression is evaluated.

[4]
8
a(i)

Define the following asymmetric key encryption terms: cipher text

[1]
a(ii)

public key.

[1]
b

Outline the process that would allow an individual to use asymmetric key encryption to send a private message to an organisation.

[4]
9
Figure 1 of 1 for question 9
a

Explain how supervised learning and unsupervised learning are trained.

[4]
b

Calculate the best path between the Begin node and each of the nodes in the diagram using Dijkstra’s algorithm. Show your working and write your answers in the table provided.

[5]
10
a

An insertion sort algorithm is used to sort the unique ID numbers of club members into ascending order. These unique ID numbers are kept in a 1D array MemberID. The corresponding names of the members are stored in a 1D array Names. The arrays are declared as: DECLARE MemberID : ARRAY[0:499] OF INTEGER DECLARE Names : ARRAY[0:499] OF STRING The following table defines the variables to be used in this algorithm. identifier | data type | description Count | integer | main loop counter to iterate through the whole array Temp1 | integer | temporary holding location for current member ID Temp2 | string | temporary holding location for current member name Insert | integer | counter to show current insertion position Write the pseudocode for an insertion sort to arrange the unique ID numbers into ascending order. Make sure that the data in both arrays is swapped as required so that data with the same index continues to apply to the same person. Use the given identifiers. Assume all variables have been declared.

[5]
b

Describe the factors, other than changing the algorithm, that affect the time efficiency of the insertion sort algorithm.

[3]
c

State the Big O notation for the worst‑case time complexity of the insertion sort.

[1]
11

A declarative programming language is used to show which home cinema devices customers already have and which devices they want. 01 device(projector). 02 device(large_tv). 03 device(medium_tv). 04 device(bluray_player). 05 device(amplifier). 06 device(media_player). 07 device(main_speakers). 08 device(centre_speaker). 09 device(surround_speakers). 10 customer(petrov). 11 customer(margaret). 12 customer(steffan). 13 customer(susan). 14 customer(allan). 15 customer(sammy). 16 owns(projector, petrov). 17 owns(media_player, petrov). 18 owns(projector, susan). 19 owns(projector, sammy). 20 owns(media_player, allan). 21 owns(bluray_player, steffan). 22 wants(main_speakers, margaret). 23 wants(surround_speakers, susan). 24 wants(medium_tv, steffan). These facts (clauses) have the meanings: fact | meaning 01 | A projector is a device. 10 | Petrov is a customer. 16 | Petrov owns a projector. 22 | Margaret wants main speakers.

a

Betty is a customer who owns a large TV and wants a Blu‑ray player. Write additional facts to represent this information. 25 ............................................................................................................................................. 26 ............................................................................................................................................. 27 .............................................................................................................................................

[3]
b

Using the variable Owner, the goal: owns(media_player, Owner) returns Owner = petrov, allan Write the result of using the goal: owns(projector, Owner) Owner = ...........................................................................................................................

[1]
c

D may be available for A to buy if D is a device and A is a customer and A does not own that device. Write this as a rule: may_buy_device(D, A) IF ............................................................................................................................................ ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................

[4]

Mark scheme

9618/31 · May/June 2026

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1a · 2 marks

  • 1 mark

    set

  • 1 mark

    class / object

1b(i) · 1 mark

  • 1 mark

    A user-defined data type that is used to reference a memory location

1b(ii) · 2 marks

  • 1 mark

    MP1 TYPE Word

  • 1 mark

    MP2 = ^STRING

2a · 1 mark

  • 1 mark

    Booking systems e.g. airline / Operating system file management systems / Video games / Encryption utilities / Compression utilities

2b · 2 marks

  • 1 mark

    MP1 Records are stored in no particular order / any available position

  • 1 mark

    MP2 A record key is used to determine the location where the record is to be stored

  • 1 mark

    MP3 … which is generated using a hash function on the key field

  • 1 mark

    MP4 New records are added in an empty location

2c · 3 marks

  • 1 mark

    MP1 Direct file access is used

  • 1 mark

    MP2 A hashing function is used on the key field to find the location in the file where the record is stored

  • 1 mark

    MP3 If the required record is found at the hashed location it is retrieved

  • 1 mark

    MP4 If the required record is not found, a collision resolution strategy is employed // Records from that point are searched linearly until the required record is found // Linear probing

3a · 3 marks

  • 1 mark

    MP1 correct method to find the binary number

  • 1 mark

    MP2 correct use of exponent

  • 1 mark

    MP3 correct answer in the space provided

3b · 3 marks

  • 1 mark

    correct application of exponent seen

  • 1 mark

    correct method to find the final answer

  • 1 mark

    Denary value: −94.25 // −94 1/4

4 · 4 marks

  • 1 mark

    Application: contains all the programs that exchange data and includes protocols: HTTP, FTP and POP3

  • 1 mark

    Transport: ensures packets arrive in sequence without errors

  • 1 mark

    Internet: identifies the intended network and host.

  • 1 mark

    Link: interconnects nodes or hosts in a network, encapsulates IP packets into frames and ensures correct protocols are followed

5 · 4 marks

  • 1 mark

    MP1 Virtual machines can provide an instruction set architecture that is different to the actual computer. // Multiple operating system environments can be used on the same computer. // You can run an application / legacy software which is not compatible with the host computer / operating system using a guest operating system that is compatible with the application / software.

  • 1 mark

    MP2 Costs can be saved by not having to purchase additional hardware to test / use applications.

  • 1 mark

    MP3 The guest operating system can crash without affecting the host machine. // Guest operating system hosted on a virtual machine can be used without impacting anything outside the virtual machine. // When trying a piece of suspicious software, if it causes a problem, it will only infect the virtual machine.

  • 1 mark

    MP4 Degraded performance. // A virtual machine is less efficient than real machines because they access hardware indirectly. // The virtual machine has to request access to the hardware from the host, which slows usability. // If multiple virtual machines are running on the same host, performance may be negatively affected if the host lacks sufficient power to deal with all their requirements.

  • 1 mark

    MP5 A virtual machine is dependent on the features / standard of the host machine. // It may be affected by the weaknesses of the host machine // Not all hardware/software can be emulated.

  • 1 mark

    MP6 Managing and maintaining a virtual machine can be quite complex /specialised.

6a · 3 marks

  • 1 mark

    One mark for columns P, Q and R

  • 1 mark

    One mark for column S

  • 1 mark

    One mark for column Z

6b · 1 mark

  • 1 mark

    Solving logic expressions using a K-map makes it easier because you don’t need to use Boolean algebra. // It is guaranteed to give you an optimal solution. // Converts complex logic problems into simpler ones.

6c(i) · 3 marks

  • 1 mark

    Loop for CD=11 (row of 1s: (00,11), (01,11), (11,11), (10,11))

  • 1 mark

    Loop for AB=11 (column of 1s: (11,00), (11,01), (11,11), (11,10))

  • 1 mark

    2x2 loop for (01,01), (11,01), (01,11), (11,11)

6c(ii) · 2 marks

  • 1 mark

    MP1 One correct Boolean term with an OR (+) sign

  • 1 mark

    MP2 All three correct Boolean terms connected by OR signs and no other terms present.

7a · 2 marks

  • 1 mark

    MP1 21 15 3 –

  • 1 mark

    MP2 8 + * 7 /

7b · 3 marks

  • 1 mark

    MP1 (c + d)

  • 1 mark

    MP2 (a - b) * (a - c)

  • 1 mark

    MP3 /( MP2 )

7c · 4 marks

  • 1 mark

    Ring 1 (stacks 1 to 3: 15, then 15 & 9, then 6)

  • 1 mark

    Ring 2 (stacks 4 to 6: 6 & 24, then 6 & 24 & 3, then 6 & 8)

  • 1 mark

    Ring 3 (stacks 7 to 10: 48, then 48 & 18, then 48 & 18 & 6, then 48 & 24)

  • 1 mark

    One mark for the interim total (48) and the final total (2).

8a(i) · 1 mark

  • 1 mark

    Text encrypted from plain text using an encryption algorithm // The result of putting plain text through an encryption algorithm.

8a(ii) · 1 mark

  • 1 mark

    A cryptographic/encryption key that can be obtained/used by anyone to send encrypted messages. // A key that is part of a matched pair // An encryption key known to all users // A decryption key obtained/used by anyone to decrypt sender’s digital signature

8b · 4 marks

  • 1 mark

    MP1 The organisation generates a matched pair of encryption keys / MP6 The individual generates a matched pair of encryption keys

  • 1 mark

    MP2 The organisation makes its public key available to the individual / MP7 The individual sends their public key to the organisation

  • 1 mark

    MP3 The individual uses the public key to encrypt their message / MP8 The individual uses their private key to encrypt their message

  • 1 mark

    MP4/MP9 … and sends this message to the organisation / MP5 The organisation decrypts the message using their private key / MP10 The organisation decrypts the message using the individual’s public key

9a · 4 marks

  • 1 mark

    MP1 Supervised learning makes use of labelled training data

  • 1 mark

    MP2 … to predict future events based on past outcomes

  • 1 mark

    MP3 Unsupervised learning makes use of unlabelled training data / training data without instructions on what to do with it / does not use training data with the correct/expected outcomes

  • 1 mark

    MP4 … to search for hidden patterns/relationships/structures/similarities/ outliers within the data

  • 1 mark

    MP5 … to therefore get closer to the desired outcome as the training progresses.

9b · 5 marks

  • 1 mark

    Working MP1 / MP2 / MP3 / MP4 / MP5 / MP6 (Max 3)

  • 1 mark

    Working MP1 / MP2 / MP3 / MP4 / MP5 / MP6 (Max 3)

  • 1 mark

    Working MP1 / MP2 / MP3 / MP4 / MP5 / MP6 (Max 3)

  • 1 mark

    Correct values: four or five correct values (1 mark)

  • 1 mark

    Correct values: all six correct values (1 additional mark)

10a · 5 marks

  • 1 mark

    MP1 Correct outer loop iterating through entire array starting at second element

  • 1 mark

    MP2 Temporary variables both initialised to current value of arrays

  • 1 mark

    MP3 Insertion variable initialised

  • 1 mark

    MP4 Correct inner loop criteria

  • 1 mark

    MP5 Elements of both arrays inserted as appropriate / MP6 Insertion variable decremented / MP7 Temporary values replaced in correct place in arrays

10b · 3 marks

  • 1 mark

    MP1 Array size // Number of elements in the array

  • 1 mark

    MP2 … time efficiency will improve if the number of elements to be sorted are fewer

  • 1 mark

    MP3 Initial order

  • 1 mark

    MP4 … if the array is partially sorted the insertion sort works better

10c · 1 mark

  • 1 mark

    O(n2) // O(N2)

11a · 3 marks

  • 1 mark

    (25) customer(betty).

  • 1 mark

    (26) owns(large_tv, betty).

  • 1 mark

    (27) wants(bluray_player, betty).

11b · 1 mark

  • 1 mark

    (Owner =) petrov, susan, sammy

11c · 4 marks

  • 1 mark

    MP1 device(D)

  • 1 mark

    MP2 customer(A)

  • 1 mark

    MP3 owns(D, A).

  • 1 mark

    MP4 All logic conditions / punctuation correct (accept , for AND)

Questions

  1. 1
    Question 15 marks
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  2. 2
    Question 26 marks
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  3. 3
    Question 36 marks
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  4. 4
    Question 44 marks
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  5. 5
    Question 54 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 79 marks
    Mark scheme ›
  8. 8
    Question 86 marks
    Mark scheme ›
  9. 9
    Question 99 marks
    Mark scheme ›
  10. 10
    Question 109 marks
    Mark scheme ›
  11. 11
    Question 118 marks
    Mark scheme ›

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